A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that increases from 0 to 0.0418 N. The length and radius of the collagen are, respectively, 2.69 cm and 0.0934 cm, and Young's modulus is 3.1 x 106 N/m2.
(a) If the stretching obeys Hooke's law, what is the spring constant k for collagen?
(b) How much work is done by the variable force that stretches the collagen?
1
Expert's answer
2010-11-16T05:47:41-0500
According Hooke's law: F= - kΔx = (EA0/L0)Δx = Δx k = (E*πR2 / L0) = (3.1x106* 3.1415* (0.0934x10-2)2 / 2.69x10-2) = 315.82 N/m E is the Young's modulus πR2 is the original cross-sectional area through which the force is applied; L0 is the original length of the object. Δx = F / k = 0.0418 / 315.18 = 0.00013 m = 1.3x10-4 mA = kx2/2 = 315.18* (1.3x10-4 )2/2 = 2.66 x 10-6 J.
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
Comments
Leave a comment